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I quote Øksendal (2003).

Let $\mathcal{V}=\mathcal{V}(S,T)$ be the class of functions $f(t,\omega):[0,\infty)\times\Omega\to\mathbb{R}$ such that $(t,\omega)\to f(t,\omega)$ is $\mathcal{B}\times\mathcal{F}$-measurable (where $\mathcal{B}$ denotes the Borel $\sigma$-algebra on $[0,\infty)$), $f(t,\omega)$ is $\mathcal{F}_t$-adapted and $\mathbb{E}\bigg[\int_{S}^T f(t,\omega)^2 dt\bigg]<\infty$.
[...] Recall that a function $\phi\in\mathcal{V}$ is called elementary if it has the form $$\phi(t,\omega)=\sum_j e_j(\omega)\cdot\chi_{[t_j, t_{j+1}]}(t)\tag{1}$$ [...]

Statement Let $g\in\mathcal{V}$ be bounded and $g(\cdot,\omega)$ continuous for each $\omega$. Then there exists elementary functions $\phi_n\in\mathcal{V}$ such that $$\mathbb{E}\left[\int_S^T\left(g-\phi_n\right)^2dt\right]\to 0\hspace{1.5cm}\text{as }n\to\infty\tag{2}$$
Proof Define $\phi_n(t,\omega)=\sum_j g(t_j,\omega)\cdot\chi_{[t_j,t_{j+1})}(t)$. Then, $\phi_n$ is elementary since $g\in\mathcal{V}$, and $$\int_S^T(g-\phi_n)^2dt\to0\hspace{1.5cm}\text{as }n\to\infty\text{ for each }\omega$$ since $g(\cdot,\omega)$ is continuous for each $\omega$. Hence $\mathbb{E}\left[\int_S^T(g-\phi_n)^2dt\right]\to0$ as $n\to\infty$ by bounded convergence.


My questions:

  1. Why does definition of continuity of $g(\cdot,\omega)$ imply that $$\displaystyle{\int_S^T(g-\phi_n)^2dt}\to0\hspace{1.5cm}\text{as }n\to\infty\text{ for each }\omega\hspace{3.5cm}\text{?}$$


My interpretation: I think I am allowed to conceive $\phi_n$ as a kind of step-function, whose value at time $t_n$ corresponds to the value of the continous and bounded function $g$ at time $t_n$. Does that mean that if I shrink the differential of time $[t_j,t_{j+1})$, continuity of $g$ implies that $|g-\phi_n|<\varepsilon\text{ for }|t_j-t_{j-1}|<\delta$ (which implies that $\displaystyle{\int_S^T(g-\phi_n)^2dt}\to0\hspace{0.5cm}\text{as }n\to\infty\text{ for each }\omega$)?


  1. In the end, is Lebesgue's dominated convergence theorem applied? If so, why does it lead from $$\displaystyle{\int_S^T(g-\phi_n)^2dt}\to0\hspace{1cm}\text{as }n\to\infty\text{ for each }\omega$$ to $$\mathbb{E}\left[\displaystyle{\int_S^T(g-\phi_n)^2dt}\right]\to0\hspace{2.3cm}\text{as }n\to\infty\text{ for each }\omega\hspace{1.8cm}\text{ ?}$$


My interpretation: What I think is that one could set $X_n=(t_{j+1}-t_j)$ and $Y_n=\displaystyle{\int_S^T(g-\phi_n(t,\omega))^2}$, which - as seen in my interpretation in point $1.$ - since $g$ is continuous, by definition of continuity, is such that for every $t$, $|Y_n|<\epsilon$ whenever $|X_n|<\delta$. In other words, $$g=\lim_{n\to\infty}\phi_n(t,\omega)\hspace{0.5cm}\text{ pointwise}\tag{3}$$ implies that $$0=\lim_{n\to\infty}\int_S^T(g-\phi_n(t,\omega))^2dt\hspace{0.5cm}\text{ pointwise}\tag{4}$$ Hence, given the immediately above explained conditions:

  • $|Y_n|<\epsilon\text{ for every }t$ (namely, "boundedness"), whenever $|X_n|<\delta$;
  • $0=\lim\limits_{n\to\infty}\displaystyle{\int_S^T(g-\phi_n(t,\omega))^2dt}\hspace{0.5cm}\text{ pointwise}$ (namely, "pointwise convergence")
    one could apply Lebesgue's dominated convergence theorem: $$\lim_{n\to\infty}\mathbb{E}\left(Y_n\right)=\mathbb{E}\left(\lim_{n\to\infty}Y_n\right)=\mathbb{E}\left(0\right)=0$$



Are my interpretations of points $1.$ and $2.$ correct? If not, why?

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  1. Yes, your reasoning is correct.
    For a fixed $\omega$, $g(t,\omega)$ is a continuous function of $t$. This means it is uniformly continuous (in $t$) on the compact interval $[S,T]$. Hence, for any given $\epsilon > 0$, we can find $\delta >0$ such that
    $$ |g(s, \omega) - g(t,\omega)| < \epsilon \quad \forall s, t \in [S,T]: |s-t| < \delta. $$ Now choose a $\epsilon > 0$ as above, and a time spacing $$ S = t_0 < t_1 \ldots < t_n = T, $$ where max $|t_i - t_{i+1}| < \delta$.
    On the every interval $[t_i, t_{i+1})$ we have
    $$ |g(t, \omega) - \phi_n(t,\omega)| = |g(t, \omega) - \phi_n(t_i,\omega)| = |g(t, \omega) - g(t_i,\omega)|< \epsilon. $$ So $$ \int_S^T |g(t, \omega) - \phi_n(t,\omega)|^2 \,dt < \epsilon^2 \cdot (T-S). $$ Since epsilon was arbitrary, the integral can be made arbitrarily small by making the maximum grid spacing, max $|t_i - t_{i+1}|$, small enough.
    You don't write what the $n$ in your functions $\phi_n$ stands for, but I assume it means the grid spacing goes to zero when $n \to \infty$.

  2. Yes, here the Lebesgue dominated convergence theorem can be applied.
    First you have to check that the integrand is uniformly bounded by an integrable function for all $n$. This follows from the previous point (EDIT: This is wrong, as forgottenarow below points out. You als need to use the boundedness of $g$ here. The subtle point here is that the n typically depends on $\omega$).

The integrand can be made smaller than $\epsilon^2$ if the grid is fine enough. This is of course an integrable function, since we consider a finite interval.
Also, for each $\omega$, you have pointwise convergence of $\phi_n(t, \omega)$ to $g(t, \omega)$ when $n\to\infty$ according to (1) (you even have uniform convergence).
So the conditions for the LDK theorem is fulfilled and you are allowed to put the limit inside the integral. So $$ \lim_{n \to \infty}\int_S^T |g(t, \omega) - \phi_n(t,\omega)|^2 \,dt = \int_S^T \lim_{n \to \infty} |g(t, \omega) - \phi_n(t,\omega)|^2 \,dt = \int_S^T 0 dt = 0. $$

Regarding the expectation, you can put the limit inside by the same type of reasoning:
Let $\mathbb{P}$ be the probability measure in which we take the expectation. $$ \begin{eqnarray*} \mathbb{E} \left[ \int_S^T |g(t, \omega) - \phi_n(t,\omega)|^2 \,dt \right] = \int \left( \int_S^T |g(t, \omega) - \phi_n(t,\omega)|^2 \,dt \right) d \mathbb{P}(\omega) \end{eqnarray*} $$ For any $\epsilon$ and $n$ large enough the inner integrand is bounded by $\epsilon^2$ according to 1. And, as I mentioned above, this means it is uniformly bounded for all $n$ by the integrable function $\epsilon^2$.
This function is integrable since $$ \int \left( \int_S^T \epsilon^2 \,dt \right) d \mathbb{P}(\omega) = (T-S)\epsilon^2 \cdot \int d\mathbb{P(\omega)} = (T-S)\epsilon^2, $$ since the total mass of a probability measure is $1$.
Hence the conditions for LDK is fulfilled and you can put the limit inside the double integral and get $0$ in the limit as $n \to \infty$ as before.

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  • $\begingroup$ Part 1 works, but I don't think we can use the same reasoning in part 2. For every $\omega$, there exists an $n$ large enough so that the bound in part 1 holds. However, this $n$ depends on $\omega$ so it can be viewed as a random variable. This also means that for any $n$, there could be a positive probability that the integral is arbitrarily large, which would prevent the integral from converging in expectation. I think we need to use the boundedness of $g$ and $\phi_n$ here to fully justify this step. $\endgroup$ – forgottenarrow Oct 26 '20 at 18:45
  • $\begingroup$ Yes, you are right. I forgot that the n depends on the $\omega$. $\endgroup$ – Jesper Tidblom Oct 26 '20 at 18:48
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Note that $g$ is assumed to be bounded, so there exists some $M < \infty $ such that $\sup_{t \in [S,T]} |g(t,\omega)| < M$ for almost surely all $\omega$. Furthermore, by definition of $\phi_n$, $\sup_{n \in \mathbb{N}}\sup_{t \in [S,T]} |\phi_n(t,\omega)| < M.$

  1. The almost sure continuity of $g$ ensures that $|g(t,\omega) - \phi_n(t,\omega)|^2 \to 0$ for all $t$ for almost surely all $\omega$. By boundedness, $\sup_t |g(t,\omega) - \phi_n(t,\omega)|^2 \leq 4M^2 < \infty$. So by bounded convergence,

$$\int_S^T|g(t,\omega) - \phi_n(t,\omega)|^2\,dt \to 0 \text{ for almost surely all } \omega\in \Omega. $$

  1. Since $\sup_t |g(t,\omega) - \phi_n(t,\omega)|^2 \leq 4M^2$ almost surely, it follows that,

$$\int_S^T|g(t,\omega) - \phi_n(t,\omega)|^2\,dt < \int_S^T 4M^2\,dt = 4M^2(T-S)<\infty \text{ for almost surely all } \omega \in \Omega.$$

So we can apply bounded convergence again to get,

$$\lim_{n\to\infty}\mathbb{E}\left[\int_S^T|g(t,\omega) - \phi_n(t,\omega)|^2\,dt\right] = \mathbb{E}\left[\lim_{n\to\infty}\int_S^T|g(t,\omega) - \phi_n(t,\omega)|^2\,dt\right] = 0.$$

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  • $\begingroup$ I have some questions: could you please detail more the reasons why: 1. "almost sure continuity of g ensures that $|g(t,\omega) - \phi_n(t,\omega)|^2 \to 0$ [...]"?; 2. "By boundedness, $\sup_t |g(t,\omega) - \phi_n(t,\omega)|^2 \leq 4M^2$". Thank you a lot in advance $\endgroup$ – Strictly_increasing Oct 30 '20 at 9:26
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    $\begingroup$ Sure! Let $\{t_i^n\}$ be the mesh which generates $\phi_n$. Let $\tau_n = \max\{t^n_i:t^n_i \leq t\}$. Recall that $\phi_n(t) = g(\tau_n)$. However, since the mesh size (maximum value of $|t^n_{i+1}-t^n_i|$) goes to zero with $n$, it follows that $\tau_n \to t$. Thus, $\lim_{n\to\infty} \phi_n(t,\omega) = \lim_{n\to\infty} g(\tau_n,\omega) = g(t,\omega)$ by a.s. continuity of $g$. For your second question, since $g$ and $\phi_n$ are both bounded by $M$, $$|g - \phi_n|^2 \leq \left(|g| + |\phi_n|\right)^2 \leq (M+M)^2 = 4M^2.$$ $\endgroup$ – forgottenarrow Oct 30 '20 at 18:38

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